Question

How to demonstrate that f is a bijective function. $ f:{{M}_{2}}\left( \mathbb{R} \right)\to {{M}_{2}}\left( \mathbb{R} \right);f\left( X \right)=AX{{A}^{-1}};A\in {{M}_{2}}\left( \mathbb{R} \right);A$- inversible?

Answer 1

Hint: In the question we are asked to demonstrate that f is a bijective function . In order to do this question, first we have to take the inverse of f(x) as h(x). Then we have to substitute x with the f(x). Then we have to simplify the equation. Then we have to substitute the x in f(x) with h(x). After solving this, if in both the equations, you get the answer as x, then it is a bijective function.

Complete step by step solution:
In the given question we have to prove that f is a bijective function,
To solve this, we have to take the inverse of f(x) as h(x). Then we have to substitute x with the f(x). Then we have to simplify the equation. Then we have to substitute the x in f(x) with g(x). After solving this, if in both the equations, you get the answer as x, then it is a bijective function.
First we take h(x) as the inverse of f. Therefore, we get:
$ \Rightarrow h\left( x \right)={{f}^{-1}}\left( x \right)={{\left( AX{{A}^{-1}} \right)}^{-1}}={{A}^{-1}}XA$
Then we will substitute h(x) in f.
$ \Rightarrow f\left( h\left( x \right) \right)=Ah\left( x \right){{A}^{-1}}=A{{A}^{-1}}XA{{A}^{-1}}$
But we know that $ A{{A}^{-1}}=I$. Therefore, substituting this we get:
$ \Rightarrow f\left( g\left( x \right) \right)=IXI=X$
Then we will substitute f(x) in h.
$ \Rightarrow h\left( f\left( x \right) \right)={{A}^{-1}}f\left( x \right)A={{A}^{-1}}AX{{A}^{-1}}A$
But we know that $ {{A}^{-1}}A=I$. Therefore, substituting this we get:
$ \Rightarrow h\left( f\left( x \right) \right)=IXI=X$

Therefore, h(x) is the inverse of f(x) and the ranges of both of them are the same. Therefore f is a bijective function.

Note: In this question, you have to know some basic identities like the $ {{A}^{-1}}A=I$. Without this you will not be able to prove the question. You also have to know the basic definition of the bijective function. Also in this question, even h(X) is a bijective function.